2 Answers
2

You've got most of the pieces you need: we also use that the union of all left cosets of H in G IS G. Now we just put the pieces together:

Let $H\le G$, $|G| = n$, and let $|H| = m$.

Since every coset (left or right) of a subgroup $H\le G$ has the same number of elements as $H$, we know that every coset of $H$ also has $m$ elements. Let $r$ be the number of cells in the partition of G into left cosets of $H$ (because the union of the left cosets of $H$ in $G$ is $G$, and these cosets are disjoint, they partition $G$).

Maybe it helps to add that the union of all the left cosets of H in G is again G. So G is the disjoint union of [G:H] sets that all have the same number of elements, namely $|H|$. Thus $|G| = |H|[G:H]$